Four charged particles namely a proton, an electron, a deuteron, a lithium ion (73Li+) enter a uniform magnetic field with the same velocity normal to its direction. Arrange them in the order of increasing frequency of motion (ν) in a circular path for the statements below:
(A) ν1 is the frequency of circular motion of proton
(B) ν2 is the frequency of circular motion of electron
(C) ν3 is the frequency of circular motion of deuteron
(D) ν4 is the frequency of circular motion of 73Li+
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (D), (C), (A), (B)
A charged particle moving perpendicular to a uniform field moves in a circle, with the magnetic force providing the centripetal force. Its frequency of revolution (cyclotron frequency) does not depend on its speed:
\( \displaystyle qvB = \frac{mv^{2}}{r} \Rightarrow r = \frac{mv}{qB} \)
\( \displaystyle \nu = \frac{v}{2\pi r} = \frac{qB}{2\pi m} \)
So \( \nu \propto q/m \). Taking the proton’s charge e and mass \( m_p \) as units:
(A) ν1 is the frequency of circular motion of proton — \( q/m = 1 \)
(B) ν2 is the frequency of circular motion of electron — \( q/m \approx 1836 \)
(C) ν3 is the frequency of circular motion of deuteron (1 proton + 1 neutron) — \( q/m = 1/2 \)
(D) ν4 is the frequency of circular motion of \( {}^{7}_{3}\mathrm{Li}^{+} \) (charge +e, mass 7mp) — \( q/m = 1/7 \)
Increasing order of frequency: \( \nu_4 \lt \nu_3 \lt \nu_1 \lt \nu_2 \), that is (D), (C), (A), (B).
NCERT (2026-27):
Motion in a Magnetic Field – “… If \( \omega \) is the angular frequency, then \( v = \omega r \). So,
\( \displaystyle \omega = 2\pi \nu = q B / m \qquad [4.6(a)] \)
which is independent of the velocity or energy. Here \( \nu \) is the frequency of rotation.”